MHB Deriving a Vector Expression from Given Direction

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To derive a vector expression from the direction defined by dx=2dy=-2dz, one must first express the relationships between the variables. This can be done by setting a parameter, such as t, to represent the changes in x, y, and z based on the given ratios. The directional derivative of a scalar function can then be calculated using the gradient and the unit vector derived from these relationships. Understanding how to convert the directional ratios into a vector representation is crucial for solving the problem. Seeking clarification on these steps is essential for progressing in the calculation of the directional derivative.
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How can you derive an expression from the given direction defined by dx=2dy=-2dz. This is just a part of a problem that I'am solving. I have no idea how to get a vector expression using the given info. please help. Thanks!

by the way I'm trying to find the directional derivative of a certain scalar function in the direction given above. By I don't know how to get a vector representayion of the given direction. Please help me!
 
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is that possible? use derivative.
 
What? I Know that I would use partial derivative here. But what I need help with is finding the vector representation of the given direction. Please anybody knows how to do it tell me. Thanks.!
 
Hey guys, do you have any idea on how to do this? I want to know it so badly. Please give me an insight.
 

Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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